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For every $4\times 4$ real non singular symmetric matrix $A$ there exist a positive integer $p$ such that

  1. $pI+A$ is positive definite.

  2. $A^p$ is positive definite.

  3. $A^{-p}$ is positive definite.

  4. $ e^{pA}-I$ is positive definite.

$ 1$st option is correct as we can choose $p$ which is greater than each eigen value of matrix $A$. Similarly option second and third is also correct as in these case choose $p$ an even integer. But I don’t know how to deal with third option . Please help me in this case . Thanks .

neelkanth
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2 Answers2

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By definition, $$e^A = \sum_{k=0}^{\infty}\frac{A^k}{k!}$$ Let $\lambda$ be an eigenvalue of $A,$ and $X$ an associated eigenvector. Then it's easy to see that $X$ is and eigenvector of $e^A$ with eigenvalue $e^\lambda.$

That is, the eigenvalues of $e^A$ are the exponentials of the eigenvalues of $A$.

Berci
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saulspatz
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If $A$ is symmetric then $A^{-1}$ is also symmetric so you reduced the third option to the second. Namely, for every even $p \in \mathbb{N}$ we have that $A^{-p}$ is positive.

For the fourth option, since $A$ is symmetric there exists an orthogonal matrix $U$ and a diagonal matrix $D$ such that $$A = U^TDU$$

Now we have

$$e^{pA} = e^{pU^TDU} = U^Te^{pD}U$$

Namely, if $D = \pmatrix{\lambda_1 & 0 &\cdots &0 \\ 0 & \lambda_2 & \cdots & 0\\ \vdots & \vdots & \ddots & \cdots \\ 0 & 0 & \cdots &\lambda_n}$ then $e^{pD} = \pmatrix{e^{p\lambda_1} & 0 &\cdots &0 \\ 0 & e^{p\lambda_2} & \cdots & 0\\ \vdots & \vdots & \ddots & \cdots \\ 0 & 0 & \cdots &e^{p\lambda_n}}$.

So $e^{pA}$ is also symmetric and with real eigenvalues $e^{p\lambda}$ where $\lambda \in \sigma(A)$.

Now $e^{pA} - I$ has eigenvalues $e^{p\lambda} - 1$.

If we take an example like $$A = \pmatrix{1 & 0 \\ 0 & -1} \implies e^{pA} - I = \pmatrix{e^p-1 & 0 \\ 0 & e^{-p}-1}$$

there is no way both $e^p-1$ and $e^{-p}-1$ can be positive so option $4$ is false.

mechanodroid
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